# Show ring has no divisors of zero

1. Jan 13, 2009

### samkolb

1. The problem statement, all variables and given/known data
Let R be a ring containing two or more elements such that for each nonzero a in R, there exists a unique b in R such that a=aba. Show that R contains no divisors of zero.

2. Relevant equations
The ring axioms.

3. The attempt at a solution
I assumed that R contains divisors of zero. So assume there exist nonzero a,b in R such that ab=0. Then there exist unique c,d in R such that a=aca and b=bdb.

Consider the expression ca+bd.

If ca+bd=0, then multiplying on the left by a gives a=0 and multiplying on the right by b gives b=0.

So assume ca+bd is not equal to zero. Then there exists unique e in R such that
ca+bd=(ca+bd)e(ca+bd).

Multiplying on the left by a gives a=aebd+aeca=ae(ca+bd).

Substituting this expression for a into a=aca gives
a=aca=ae(ca+bd)ca.

From the uniqueness of c, e(ca+bd)c=c.

This is as far as I've gotten. I don't know if I have made any progress or am just going in circles.

2. Jan 13, 2009

### Dick

Suppose a and c are zero divisors. Then ac=0. There is a UNIQUE b such that aba=a. What is a(b+c)a?